GCSE maths coursework Essay Example

GCSE maths coursework Essay For this piece of GCSE maths coursework I have been asked to investigate the relationship between the total of a three-step stair on a number grid and its position on the grid. The stair will be on a 1010 number grid like this:919293949596979899100818283848586878889907172737475767778798061626364656667686970515253545556575859604142434445464748495031323334353637383940212223242526272829301112131415161718192012345678910I would need to find the total for stairs on the grid. For example the one shaded in red:45+35+25+36+26+27=194I would then tabulate this result and move the stair one square to the right so then I would get.46+36+26+37+27+28=200I would then continue moving it to the right until I am unable to anymore.Part 1:I will start of in the bottom left hand corner and work myself to the right of the grid.Position 1211112123The total for this stair is:21+11+1+12+2+3=50Position 2221213234The total for this stair is:22+12+2+13+3+4=56Position 3231314345The total for this stair is:23+13+3 +14+4+5=62I will now tabulate these results and others.Position Number (Lowest number in grid)12345678Total5056626874808692From looking at my results I can see that when I move the stair shape left by 1 column the total increases by 6 each time. I am going to try to find a formula to explain the relationship between the position and the total. For the rest of my investigation when I refer to n it will be the lowest number in the stair (The number in the bottom left hand corner). If you look at:25+26+27+35+36+45=194You can get the following formula:n+(n+1)+(n+2)+(n+10)+(n+11)+(n+20)=194Which if you simplify gives you a general formula of:6n+44=TotalI can prove this formula works by looking at my table once more. And adding some more to itPosition Number (Lowest number in grid)12345678Total5056626874808692DifferenceBetween Totals66666666As I mentioned above the difference between each stair total is 6. This shows that 6n is somewhere in the formula. I will now see what relation 6n has to the stair totals.6n612182430364248Total5056626874808692The value for 6n and the stair total are different. The total is larger which means the formula must be 6n+?. I will see what the difference between the total and 6n is to get the value of ?:Difference4444444444444444You have to add 44 to 6n to get the total of the stair. So now you can see the formula for the three-step stair is 6n+44 like I stated above.This formula should work for obvious reasons. If you look at position 1 the biggest number is 21 and the smallest 1, meaning there is a difference of 20 between them. This will be the same anywhere on the grid. In the stair below there is a difference of 20 between 94 and 74. This relationship between the numbers remains the same anywhere on the grid and that is why my formula works anywhere.I will now test to see if my formula, 6n+44, does work.948485747576The total for this stair is:74+75+76+84+85+94=488For this stair shape n=74. I will now substitute numbers into my form ula 6n+44.=(674)+44=444+48=488As you can see the answer gained from my equation is the right stair total for the stair given. I will now test for another stair to prove my formula does work54+55+56+64+65+74=368n=54=(546)=44=324+44=368I have now found and proven a general formula for a three-step stair on a 1010 grid. I will now extend my investigation further.Part 2:I will now extend my investigation by posing this question.Will my formula, 6n+44, work on different sized stair shapes on the 1010 grid? If not can a formula be deduced to find out the total for any size stair on a 1010 grid.Since I already have results for three-step stairs I will begin my investigation with a four-step stair. A four-step stair on the grid will look like this919293949596979899100818283848586878889907172737475767778798061626364656667686970515253545556575859604142434445464748495031323334353637383940212223242526272829301112131415161718192012345678910Position 1The total for this stair is:1+2+3+4+11+12+13+2 1+22+31=120Position 2The total for this stair is:2+3+4+5+12+13+14+22+23+32=130Position 3The total for this stair is:3+4+5+6+13+14+15+23+24+33=140Position 4The total for this stair is:4+5+6+7+14+15+16+24+25+34=150I will now tabulate these results and those for other 4 step stairs.Position Number1234567StairTotal120130140150160170180I will see whether 6n+44 will work to find the stair total for 4 steps. I will use position 1 as an example.n=1=6n+44=(61)+44=6+44=50As you can see the formula does not apply to four step stairs. Now I will try to find a formula that does apply to four steps.Since this time there is a difference of 10 between the totals I now times n by 10 in my equation. I will see what 10n looks like compared to the totals.10n10203040506070StairTotal120130140150160170180Difference110110110110110110110The same way as with the three steps 10n does not equal the total it is larger. The difference is 110 so the formula for four step stairs is 10n+110. I will now test my new formula.61+62+63+64+71+72+73+81+82+91=720n=61=10n+110=(1061)+110=720You can now see that my formula for the four step stairs is correct. I will now try five step stairs.Position 1The total for this stair is:1+2+3+4+5+11+12+13+14+21+22+23+31+32+41=235Position 2The total for this stair is:2+3+4+5+6+12+13+14+15+22+23+24+32+33+42=250Position 3The total for this stair is:3+4+5+6+7+13+14+15+16+23+24+25+33+34+43=265Position 4The total for this stair:4+5+6+7+8+14+15+16+17+24+25+26+34+35+44=280Position 5The total for this stair:5+6+7+8+9+15+16+17+18+25+26+27+35+36+45=295Position 6The total for this stair:6+7+8+9+10+16+17+18+19+26+27+28+36+37+46=310Position Number123456StairTotal235250265280295310For five step stairs I believe that I will have to work out a new formula. Since the totals increase by 15 each time in my equation I will have to 15n.When comparing 15n to the totals I find:15n153045607590StairTotal235250265280295310Difference220220220220220220I have to add a number to 15n to get th e stair total. From my table you can see that this number is 220.The formula to find the total of any five-step stair o a 1010 grid is:15n+220I will test my formula.n=5555+56+57+58+59+65+66+67+68+75+76+77+85+86+95=1045=15n+220=(1555)+220=825+220=1045So my formula for five-step stairs does work as shown above.I will now try to find a general formula to find any size step on a 1010 grid. To try and find a pattern I will keep n the same, n=1, and change the stair size.For now in all stairs n=1One step stairThe total is1+0=1Two step stairThe total is:1+2+11=14Three step stairThe total for this stair is:1+2+3+11+12+21=50Four step stairThe total for this stair is:1+2+3+4+11+12+13+21+22+31=120Five step stairThe total for this stair is:1+2+3+4+5+11+12+13+14+21+22+23+31+32+41=235Six step stairThe total for this stair is:1+2+3+4+5+6+11+12+13+14+15+21+22+23+24+31+32+33+41+42+51=406Seven step stairThe total for this stair is:1+2+3+4+5+6+7+11+12+13+14+15+16+21+22+23+24+25+31+32+33+34+41+42+43+51 +52+61=644I will now try to find a pattern with my results.n1111111Height of Stair1234567Total114501202354066441st Difference1336701151712382nd Difference23344556673rd Difference11111111From looking at this table I am able to recognize a new idea, that in the final formula the number of 11 and heightà ¯Ã‚ ¿Ã‚ ½.I will make a new table to try and find out more.(h=height of the square)n1111111Height of Stair1234567+11h31188297704137523763773-11h-11-22-33-44-55-66-77Total066264660132023103696h3182764125216343Stair Total11450120235406644As you can see no pattern is obvious from this table.When I was looking at the formula, 6n+44 I noticed that when you multiply it by 2 the n value is a multiple of the height of the stair. I decided to check if this was the same for the other formulas.h345Formula6n+4410n+11015n+220hx2122030hà ¯Ã‚ ¿Ã‚ ½91625The relationship is the same and I put in hà ¯Ã‚ ¿Ã‚ ½ to see if there was a pattern. From the table you can see that the third row equals hà ¯Ã‚ ¿Ã‚ ½+h. This could not be used in a formula because it twice as large so to use it in my general formula it would have to be:n(hà ¯Ã‚ ¿Ã‚ ½+h)2This is the first part of my general formula but it is not complete and I need a second half.From looking at the differences in stair total and the totals from 11hà ¯Ã‚ ¿Ã‚ ½-11h I believe I have found the second part of my equation. When looking at the total of 11hà ¯Ã‚ ¿Ã‚ ½-11h when h=2 I saw it equaled 66. The formula for a two-step square is 3n+11. 66/11=6. I wondered if dividing 11hà ¯Ã‚ ¿Ã‚ ½-11h by six on other heights would work, To check I made a table:n1111111Height of Stair1234567+11h31188297704137523763773-11h-11-22-33-44-55-66-77Total066264660132023103696Divide by 601144110220Total1n+02n+116n+4410n+11015+220So every time when 11hà ¯Ã‚ ¿Ã‚ ½-11h is divided by 6 it gives the right number in the second part of each formula. So I now have a complete general formula to find out any sized stair anywhere on a 1010 number grid. Th e formula is:n+(hà ¯Ã‚ ¿Ã‚ ½+h) + (11hà ¯Ã‚ ¿Ã‚ ½-11h)2 6However I will need to test this formula:T: Total of the stairn: Stair number (the number in the bottom left had corner of the stair)h: The height of the stairn=72 Three Step Stair:T=72+73+74+82+83+92=476T=n((hà ¯Ã‚ ¿Ã‚ ½+h)/2)+(11hà ¯Ã‚ ¿Ã‚ ½-11h)/6T=72((3à ¯Ã‚ ¿Ã‚ ½+3)/2)+(113à ¯Ã‚ ¿Ã‚ ½-113)/6T=72((12)/2)+(264)/6T=72(6)+44T=432+44T=476For this one the answer is correct now I will try another sized stairn=55 Five Step Stair:T=55+56+57+58+59+65+66+67+68+75+76+77+85+86+95=1045T=n((hà ¯Ã‚ ¿Ã‚ ½+h)/2)+(11hà ¯Ã‚ ¿Ã‚ ½-11h)/6T=55((5à ¯Ã‚ ¿Ã‚ ½+5)/2)+(115à ¯Ã‚ ¿Ã‚ ½-115)/6T=55((30)/2)+(1320)/6T=55(15)+220T=825+220T=1045Now since I have tested it twice I am sure that my formula is correct and will work for any size stair anywhere on in the 1010 grid. Now looking back I see what else I could have done to improve my investigation. Instead of changing stair height I could have instead changed the width of the grid or the heig ht. I could also have changed the width and length simultaneously i.e. 55, 66, 77. GCSE Maths Coursework Essay Example GCSE Maths Coursework Essay I will then investigate the diagonal difference of a 44 grid, 33 grid and 22 grid the numbers for these grids will be taken from the 88 grid above.The way to investigate diagonal difference is shown in the example:Sarah writes down a 33 grid from the above table.101112181920262728She notices that when you multiply the opposite corners the difference between the products is 32For example:10 x 28 = 28012 x 26 = 312The diagonal difference is 312 280 = 32.What I think will happenI think that the diagonal difference will always be 32 with a 33 grid.I think that the diagonal difference for a 22 grid will be 8 each time.I am also going to investigate the diagonal difference of a grid sized 44; I think the difference will be 72 each time.I came to these conclusions by doing a series of preliminary investigations.Preliminary InvestigationsFor a 33 grid:123910111718191 x 19 = 193 x 17 = 5151 19 = 3246474854555662636446 x 64 = 294448 x 62 = 29762976 2944 = 32For a 44 grid:2627282934353637424 344455051525326 x 53 = 137829 x 50 = 14501450 1378 = 721234910111217181920252627281 x 28 = 284 x 25 = 100100 28 = 72For a 22 grid:4445525344 x 53 = 233245 x 52 = 23402340 2332 = 83334414233 x 42 = 138634 x 41 = 13941394 1386 = 8What I will Measure/ObserveI will measure the diagonal difference from a 44 grid, a 33 grid and a 22 grid. I will take these grids from a table of 88 numbers ranging from 1-64.I will find the diagonal difference by taking the four numbers from each corner multiply them by the number in the apposite corner and then I will take away the two numbers found as a result of multiplying the four opposite diagonals. This will then leave me with my diagonal difference.Examples of the type of measurements I will be gathering are found in my preliminary investigations.How I will Carry out any MeasurementsI will measure the times value of the corners of the 44 grid, 33 grid, and a 22 grid then I will minus the two answers given for each grid and find the diagonal diff erence.How many Times I will Repeat the MeasurementsI will produce four sets of results for each of the 44 grids, 33 grids and 22 grids.I will take the numbers for my smaller grids from the original 88 grid of numbers.I will gather the numbers for the smaller grids, as If I were to place the smaller grid sizes on to the 88 grid taking the right number range for the suitable grids.Taking this amount of readings provides me with broad range of results enabling me to have a more conclusive view of the investigation and it will also help me when concluding the investigation.What I will be Altering in order to make ComparisonsI am altering the sizes of my grids from 44 to 33 and then to a 22 grid size.I can then compare these grid sizes to the original grid size of 33.How I will carry out these AlterationsI will carry out these alterations by reducing the size of my grids which I will be taking the diagonal difference from.What Range of Values will be compared?I will collect four sets of results for all of the three grid sizes.I will then compare the diagonal difference values.I will then analyse these four sets of results by putting them into tables and formatting them in charts, this ill then enable me to make comparison between the diagonal differences of each grid size.What I will keep the same to make it a Fair TestI will keep the initial 88 grid the same; this is where I will gather my numbers from for the other smaller grid sizes.The 88 grid will range from 1-64 in numerical order I will always keep this the same.What Safety Precautions will need to be taken?I will not have to take any safety precautions for this investigation as no safety hazards will occur as a result of this investigation.What graphs and calculations do I intend to do? GCSE Maths Coursework Essay Example GCSE Maths Coursework Essay I will then investigate the diagonal difference of a 44 grid, 33 grid and 22 grid the numbers for these grids will be taken from the 88 grid above.The way to investigate diagonal difference is shown in the example:Sarah writes down a 33 grid from the above table.101112181920262728She notices that when you multiply the opposite corners the difference between the products is 32For example:10 x 28 = 28012 x 26 = 312The diagonal difference is 312 280 = 32.What I think will happenI think that the diagonal difference will always be 32 with a 33 grid.I think that the diagonal difference for a 22 grid will be 8 each time.I am also going to investigate the diagonal difference of a grid sized 44; I think the difference will be 72 each time.I came to these conclusions by doing a series of preliminary investigations.Preliminary InvestigationsFor a 33 grid:123910111718191 x 19 = 193 x 17 = 5151 19 = 3246474854555662636446 x 64 = 294448 x 62 = 29762976 2944 = 32For a 44 grid:2627282934353637424 344455051525326 x 53 = 137829 x 50 = 14501450 1378 = 721234910111217181920252627281 x 28 = 284 x 25 = 100100 28 = 72For a 22 grid:4445525344 x 53 = 233245 x 52 = 23402340 2332 = 83334414233 x 42 = 138634 x 41 = 13941394 1386 = 8What I will Measure/ObserveI will measure the diagonal difference from a 44 grid, a 33 grid and a 22 grid. I will take these grids from a table of 88 numbers ranging from 1-64.I will find the diagonal difference by taking the four numbers from each corner multiply them by the number in the apposite corner and then I will take away the two numbers found as a result of multiplying the four opposite diagonals. This will then leave me with my diagonal difference.Examples of the type of measurements I will be gathering are found in my preliminary investigations.How I will Carry out any MeasurementsI will measure the times value of the corners of the 44 grid, 33 grid, and a 22 grid then I will minus the two answers given for each grid and find the diagonal diff erence.How many Times I will Repeat the MeasurementsI will produce four sets of results for each of the 44 grids, 33 grids and 22 grids.I will take the numbers for my smaller grids from the original 88 grid of numbers.I will gather the numbers for the smaller grids, as If I were to place the smaller grid sizes on to the 88 grid taking the right number range for the suitable grids.Taking this amount of readings provides me with broad range of results enabling me to have a more conclusive view of the investigation and it will also help me when concluding the investigation.What I will be Altering in order to make ComparisonsI am altering the sizes of my grids from 44 to 33 and then to a 22 grid size.I can then compare these grid sizes to the original grid size of 33.How I will carry out these AlterationsI will carry out these alterations by reducing the size of my grids which I will be taking the diagonal difference from.What Range of Values will be compared?I will collect four sets of results for all of the three grid sizes.I will then compare the diagonal difference values.I will then analyse these four sets of results by putting them into tables and formatting them in charts, this ill then enable me to make comparison between the diagonal differences of each grid size.What I will keep the same to make it a Fair TestI will keep the initial 88 grid the same; this is where I will gather my numbers from for the other smaller grid sizes.The 88 grid will range from 1-64 in numerical order I will always keep this the same.What Safety Precautions will need to be taken?I will not have to take any safety precautions for this investigation as no safety hazards will occur as a result of this investigation.What graphs and calculations do I intend to do? Gcse Maths Coursework Essay Example Gcse Maths Coursework Essay I will then investigate the diagonal difference of a 44 grid, 33 grid and 22 grid the numbers for these grids will be taken from the 88 grid above.The way to investigate diagonal difference is shown in the example:Sarah writes down a 33 grid from the above table.101112181920262728She notices that when you multiply the opposite corners the difference between the products is 32For example:10 x 28 = 28012 x 26 = 312The diagonal difference is 312 280 = 32.What I think will happenI think that the diagonal difference will always be 32 with a 33 grid.I think that the diagonal difference for a 22 grid will be 8 each time.I am also going to investigate the diagonal difference of a grid sized 44; I think the difference will be 72 each time.I came to these conclusions by doing a series of preliminary investigations.Preliminary InvestigationsFor a 33 grid:123910111718191 x 19 = 193 x 17 = 5151 19 = 3246474854555662636446 x 64 = 294448 x 62 = 29762976 2944 = 32For a 44 grid:2627282934353637424 344455051525326 x 53 = 137829 x 50 = 14501450 1378 = 721234910111217181920252627281 x 28 = 284 x 25 = 100100 28 = 72For a 22 grid:4445525344 x 53 = 233245 x 52 = 23402340 2332 = 83334414233 x 42 = 138634 x 41 = 13941394 1386 = 8What I will Measure/ObserveI will measure the diagonal difference from a 44 grid, a 33 grid and a 22 grid. I will take these grids from a table of 88 numbers ranging from 1-64.I will find the diagonal difference by taking the four numbers from each corner multiply them by the number in the apposite corner and then I will take away the two numbers found as a result of multiplying the four opposite diagonals. This will then leave me with my diagonal difference.Examples of the type of measurements I will be gathering are found in my preliminary investigations.How I will Carry out any MeasurementsI will measure the times value of the corners of the 44 grid, 33 grid, and a 22 grid then I will minus the two answers given for each grid and find the diagonal diff erence.How many Times I will Repeat the MeasurementsI will produce four sets of results for each of the 44 grids, 33 grids and 22 grids.I will take the numbers for my smaller grids from the original 88 grid of numbers.I will gather the numbers for the smaller grids, as If I were to place the smaller grid sizes on to the 88 grid taking the right number range for the suitable grids.Taking this amount of readings provides me with broad range of results enabling me to have a more conclusive view of the investigation and it will also help me when concluding the investigation.What I will be Altering in order to make ComparisonsI am altering the sizes of my grids from 44 to 33 and then to a 22 grid size.I can then compare these grid sizes to the original grid size of 33.How I will carry out these AlterationsI will carry out these alterations by reducing the size of my grids which I will be taking the diagonal difference from.What Range of Values will be compared?I will collect four sets of results for all of the three grid sizes.I will then compare the diagonal difference values.I will then analyse these four sets of results by putting them into tables and formatting them in charts, this ill then enable me to make comparison between the diagonal differences of each grid size.What I will keep the same to make it a Fair TestI will keep the initial 88 grid the same; this is where I will gather my numbers from for the other smaller grid sizes.The 88 grid will range from 1-64 in numerical order I will always keep this the same.What Safety Precautions will need to be taken?I will not have to take any safety precautions for this investigation as no safety hazards will occur as a result of this investigation.What graphs and calculations do I intend to do?